A survey was carried out to find the amount of money spent by 100 customers in a supermarket. This information was grouped as follows.

1. Identify the class intervals and their frequencies from the table.

The class intervals are all of width $25$:

$0<x\leq25,\quad 25<x\leq50,\quad 50<x\leq75,\quad 75<x\leq100,\quad 100<x\leq125,\quad 125<x\leq150.$

Their frequencies are respectively

$13,\ 23,\ 20,\ 17,\ 11,\ 16.$

2. Draw the histogram bars with correct widths and heights.

Since each class interval has the same width, use the number of customers as the bar height.

Draw adjacent bars over:

$0\text{ to }25\text{ height }13,$
$25\text{ to }50\text{ height }23,$
$50\text{ to }75\text{ height }20,$
$75\text{ to }100\text{ height }17,$
$100\text{ to }125\text{ height }11,$
$125\text{ to }150\text{ height }16.$

1. Find which class contains the median customer amount.

There are $100$ customers, so the median lies between the $50$th and $51$st values.

Cumulative frequencies:

$13,\quad 13+23=36,\quad 36+20=56$

Since the $50$th and $51$st values lie in the class where the cumulative frequency reaches $56$, the median class is

$50 < x \leq 75$.

1. Find the class interval with the highest number of customers.

The frequencies are $13, 23, 20, 17, 11, 16$.

The greatest frequency is $23$, which corresponds to the class interval $25 < x \leq 50$.

1. State one advantage of presenting the grouped survey results using a histogram.

A histogram displays grouped continuous data visually, so it is easy to compare how many customers fall in each spending interval.

1. Find the probabilities for the first coin drawn.

There are $10$ coins in total:

$4+3+3=10$

So the first branches are

$P(20\text{¢})=\frac{4}{10},\quad P(50\text{¢})=\frac{3}{10},\quad P(\$1)=\frac{3}{10}.$

2. Find the conditional probabilities for the second coin, remembering that the draw is without replacement.

After one coin is taken, there are $9$ coins left.

If the first coin is $20\text{¢}$, the remaining coins are $3$ of $20\text{¢}$, $3$ of $50\text{¢}$ and $3$ of $\char36{}1$:

$\frac{3}{9},\ \frac{3}{9},\ \frac{3}{9}.$

If the first coin is $50\text{¢}$, the remaining coins are $4$ of $20\text{¢}$, $2$ of $50\text{¢}$ and $3$ of $\char36{}1$:

$\frac{4}{9},\ \frac{2}{9},\ \frac{3}{9}.$

If the first coin is $\char36{}1$, the remaining coins are $4$ of $20\text{¢}$, $3$ of $50\text{¢}$ and $2$ of $\char36{}1$:

$\frac{4}{9},\ \frac{3}{9},\ \frac{2}{9}.$

1. Find the probability that the two coins have the same denomination.

$P(\text{same})=\frac{4}{10}\cdot\frac{3}{9}+\frac{3}{10}\cdot\frac{2}{9}+\frac{3}{10}\cdot\frac{2}{9}$
$=\frac{12+6+6}{90}=\frac{24}{90}=\frac{4}{15}$

2. Use the complement to find the probability of different denominations.

$P(\text{different})=1-P(\text{same})$
$=1-\frac{4}{15}=\frac{11}{15}$

1. Identify the outcomes where the total value is more than \$1.

The total is more than $1 for:

$20\text{¢}+\char36{}1$, $\char36{}1+20\text{¢}$, $50\text{¢}+\char36{}1$, $\char36{}1+50\text{¢}$, and $\char36{}1+\char36{}1$.

2. Add the probabilities of the favourable outcomes.